Let $f$ be an entire function. Then there exist numbers $a_0,a_1,\ldots$, independent of $z$, such that
$$f(z)=\sum_{n=0}^\infty a_n J_n(z),\quad \forall z\in\mathbb{C}$$
where $J_n$ is the Bessel function of the first kind of order $n$. This is called a *Neumann series* of $f$.

**Question**

Given $f$, are the numbers $a_n$ unique?

**Thoughts**

I took a look at *Abramowitz and Stegun* and *Whittaker and Watson*; in each book they mention the series but say nothing about the uniqueness of the coefficients. It would be nice if Neumann series turn out to be unique just as Taylor series of entire functions. I figured out that it is sufficient to prove that, for all $z\in\mathbb{C}$,

$$\sum_{n=0}^\infty a_n J_n(z)=0\implies \forall n\in\mathbb{N}:a_n=0;$$

Whittaker and Watson arrived at one possible integral representation of $a_n$ in terms of "Neumann polynomials". They essentially proved that **if** $a_n=\text{integral expression}$ **then** $f(z)=\sum_{n=0}^\infty a_n J_n(z)$, not the other way around.

*This question has also been asked on MSE as can be seen at this link*. *According to the user Conrad in the comments, the uniqueness is known if the Neumann series are required to be uniformly convergent, but not known if they're only pointwise convergent.*

**Edit:**

I'm thinking about using "orthogonality", i.e. $$\int_0^\infty \frac{J_{v+2l+1}(t)J_{v+2m+1}(t)}{t}\, dt=\frac{\delta_{l,m}}{2(2l+v+1)}$$ where $v+l+m\gt -1$ and then concluding that if $$\sum_{n=0}^\infty a_n J_n (z)=\sum_{n=0}^\infty b_n J_n(z),$$ then $$a_n=b_n$$ by the above orthogonality identity, but this requires me to justify switching $\sum_{n=0}^\infty$ with $\int_0^\infty.$

*Please note that nothing about uniform convergence is assumed in this question; suppose that Neumann series only converge pointwise, nothing more.*

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